Max wheel size according to torque

  • Thread starter Thread starter Jones1987
  • Start date Start date
  • Tags Tags
    Max Torque Wheel
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
5 replies · 5K views
Jones1987
Messages
74
Reaction score
0
Hi, I'm researching into what will be the best size wheels to put onto a motor to allow my robot to climb an incline, and then possibly find out what is the maximum incline it can reach.

I'm using this thread as a source for the math:

http://forums.trossenrobotics.com/archive/index.php/t-2900.html

But where he states this

"So, given 47 oz-in of torque with four motors (1.3 Newton-meters), mass of 2 Kg and an angle of 15 degrees, the answer is:

r = 1.3 / (9.8 * 2 * 0.26) = 0.26 m"

Would you not use the torque of a single motor to find out its max wheel size? Rather than combine the total torque and use that?

Also, once I've found my ideal size of wheel to tackle a 15degree incline, I'm taking an educated guess that a wheel to combat a 20degree incline be a smaller?
 
Physics news on Phys.org
If you assume that your robot has 4 wheels of radius r, each connected to a motor with torque T, the force balance equation is:

m*g*sin(theta) = (T/r)1 + (T/r)2 + (T/r)3 + (T/r)4 = 4*(T/r) = (4*T)/r

and 4*T is the sum of the torque from all motors. This is why you need the torque from all motors.

Don't forget that this is the maximum wheel radius for your robot to be able to climb that hill (actually it will be only just enough for preventing your robot to go downhill); you can (and should) go smaller.

They are also 2 other characteristics that will limit the climbing ability of your robot: its traction and its flip over limit.

Read this http://hpwizard.com/car-performance.html" for more info (look at the bottom of the page: Theory»Longitudinal acceleration»Accelerating»Hill climbing)
 
Last edited by a moderator:
jack action said:
If you assume that your robot has 4 wheels of radius r, each connected to a motor with torque T, the force balance equation is:

m*g*sin(theta) = (T/r)1 + (T/r)2 + (T/r)3 + (T/r)4 = 4*(T/r) = (4*T)/r

and 4*T is the sum of the torque from all motors. This is why you need the torque from all motors.

Don't forget that this is the maximum wheel radius for your robot to be able to climb that hill (actually it will be only just enough for preventing your robot to go downhill); you can (and should) go smaller.

They are also 2 other characteristics that will limit the climbing ability of your robot: its traction and its flip over limit.

Read this http://hpwizard.com/car-performance.html" for more info (look at the bottom of the page: Theory»Longitudinal acceleration»Accelerating»Hill climbing)

Hi jack, thanks for the detailed reply. I assume to obtain max wheel size, I just use sum of torque, divided by the force. Which will give me a max radius to use

So e.g.
m = 2kg
T = 0.355 * 2 (2 motors) = 0.71Nm
F = (9.8 * 2 * sin(15)) = 5.07N

So Rmax = 0.71 / 5.07
Rmax = 0.14m

Is this correct?

I'm looking at the flip over limit, and I don't know if it's because I've not long been awake, but I'm not grasping it. Will the resultant value of tan(theta) be the max angle before this robot will begin to lift and flip over?
 
Last edited by a moderator:
Yes, your calculations are correct.

As for the flip over limit, it is the angle at which the normal force on the front axle will be zero and the total weight of the vehicle will be on the rear axle. This is an unstable condition. At this point, the front axle can lift to any height and the vehicle may flip over.
 
jack action said:
Yes, your calculations are correct.

As for the flip over limit, it is the angle at which the normal force on the front axle will be zero and the total weight of the vehicle will be on the rear axle. This is an unstable condition. At this point, the front axle can lift to any height and the vehicle may flip over.

Ok brilliant. However for some reason, when I calc the ideal wheel size on a flat surface, I end up getting a lower number than what it would be on an incline. But surely on a flat surface you are allowed larger wheels?

R = T / (g * m)
R = 0.71 / (9.8 * 2)
R = 0.036m

So I have a feeling I'm missing something obvious here.